Question

If ${I_o}$ is the intensity of the principal maximum in the slit diffraction pattern, then what will be its intensity when the slit width is doubled:
(A) $2{I_o}$
(B) $4{I_o}$
(C) ${I_o}$
(D) $\dfrac{{{I_o}}}{2}$

Answer 1

In this question, we need to find out how the principal maxima varies with the slit width. We know it is inversely proportional to the slit width. Therefore, as the slit width increases, the size of principal maxima decreases. Then using this relation, we need to find the change in intensity by considering the relation between the intensity and area.

Complete Step-By-Step Solution:
We know that the width of the central maxima also known as the principal maxima is given as:
$\Delta \theta = \dfrac{{2\lambda }}{b}$
Where,
$\Delta \theta =$ Width of the principal maxima
$\lambda =$ Wavelength of light
$b =$ Slit width
From the above expression It is evident that, as slit width is doubled, the width of central maxima gets halved.
Therefore, we can write:
$\Delta \theta = \dfrac{\lambda }{b}$
As the width of the central maxima decreases, the area where the light focuses reduces and thus the intensity of light increases.
In other words we can say, intensity is inversely proportional to area, so as area increases, intensity decreases, and as area decreases intensity increases.
${I_o} \propto \dfrac{1}{A}$
Where,
${I_o}$ is the intensity and $A$ is the area.
Area varies with the square of width of the central maxima, thus we can say area becomes $\dfrac{1}{4}$.
Intensity varies, inversely with area, so intensity becomes $4{I_o}$
This is the required solution.

Hence, option (B) is correct.

Note:
Intensity of light implies the amount of energy transferred per unit area. Intensity varies as the square of amplitude, therefore as amplitude increases, the intensity of the light also increases.